Differential cross section

[Bill Foster]

Homework Statement

a) Define the differential cross section $$\frac{d\sigma}{d\Omega}$$ in “experimental” terms, i.e., in terms of the number of scattered and incident particle.

b) Show that for a situation in which there is a scattered wave $$f^+\left(\theta\right)\frac{e^{ikr}}{r}$$ asymptotically and an incident wave $$e^{i\vec{k}\dot\vec{r}}$$, the above definition corresponds to $$\frac{d\sigma}{d\Omega}=|f^+\left(\theta\right)|^2$$ . Consider that the current density associated with a wave function $$\Psi \left(\vec{r}\right)$$ is given by $$\vec{j}=-\frac{i\hbar}{2m}\left[\Psi^*\vec{\nabla}\Psi-\Psi\left(\vec{\nabla}\Psi\right)^* \right]$$.

The Attempt at a Solution

$$\frac{d\sigma}{d\Omega}=\frac{\frac{F_s}{F_i}N}{usa}$$

Where $$F_s$$ is scattered flux, $$F_s$$ is incident flux, N is unit of surface. And usa is unit of solid angle.

I also have $$I_r=I_iN\sigma$$ where $$I_r$$ is current of scattered particles, and $$I_i$$ is current of incident particles. N is unit surface.

It seems to me I need to relate N to the solid angle. But I'm probably wrong.

Some guidance here would be appreciated.

 

[mark726]

a) The differential cross section \frac{d\sigma}{d\Omega} can be defined as the measure of the probability of a particle being scattered into a particular solid angle (\Omega) from an incident beam of particles. In experimental terms, it can be written as the ratio of the number of scattered particles per unit time and per unit solid angle (dN/dt d\Omega) to the incident flux of particles per unit time and per unit area (F_i). Mathematically,

\frac{d\sigma}{d\Omega}=\frac{dN/dt d\Omega}{F_i}

b) In the given situation, the scattered wave can be written as f^+\left(\theta\right)\frac{e^{ikr}}{r}. This can be interpreted as the scattered flux (F_s) being proportional to the amplitude of the scattered wave (f^+\left(\theta\right)) and the inverse square of the distance from the source (1/r^2). Similarly, the incident wave can be written as e^{i\vec{k}\dot\vec{r}}, which can be interpreted as the incident flux (F_i) being proportional to the amplitude of the incident wave (1) and the inverse square of the distance from the source (1/r^2).

Using these interpretations, we can rewrite the definition of the differential cross section as:

\frac{d\sigma}{d\Omega}=\frac{F_s}{F_i}=\frac{|f^+\left(\theta\right)|^2}{1}=\frac{|f^+\left(\theta\right)|^2}{1/r^2}=\frac{|f^+\left(\theta\right)|^2 r^2}{1}=|f^+\left(\theta\right)|^2

This shows that the definition of the differential cross section in terms of the number of scattered and incident particles (as given in part a) is equivalent to the definition of the differential cross section in terms of the scattered wave and incident wave (as given in part b).

Now, to relate the differential cross section to the current density (j), we can use the relation:

j=\frac{dN}{dt}=\frac{dN}{d\Omega}\frac{d\Omega}{dt}=\frac{dN}{d\Omega}\frac{d\Omega}{d\phi}\frac{d\